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2 years ago

Phosphoric acid has a pka of 2.1. at what ph will 75% of phosphoric acid be in the conjugate base form?

Chemistry

2 answers:

Colt1911 [192]2 years ago

8 0

Answer:

pH is 2.58

Explanation:

The pH of the buffer made from the mixture of phosphoric acid (H₃PO₄) and its conjugate base form (H₂PO₄⁻) follows the Henderson-Hasselbalch equation:

pH = pka + log [H₂PO₄⁻] /[H₃PO₄]

If 75% of the buffer is in the conjugate base form (H₂PO₄⁻), 25% will be as H₃PO₄. Replacing to find pH:

pH = 2.1 + log [75%] /[25%]

pH = 2.58

pH is 2.58

EleoNora [17]2 years ago

5 0

solution:

$from henderson hasselbalch equation\\pH=pka+log[\frac{conjugate}{acid}]\\given[\frac{conjugate base}{acid}]\\=\frac{75}{25}=3\\PH=2.1+log3\\pH=2.1+0.5=2.6$

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A reaction container holds 5.77 g of P4 and 5.77 g of O2.

Dvinal [7]

Answer:

a) O2 is the limiting reactant

b) 5.75 grams P4O10

c) 5.79 grams P4O6

Explanation:

Step 1: Data given

Mass of P4 = 5.77 grams

Mass of O2 = 5.77 grams

Molar mass of P4 = 123.90 g/mol

Molar mass O2 = 32.0 g/mol

Step 2: The balanced equation

P4 + 3O2 → P4O6

Step 3: Calculate moles of P4

Moles P4 = mass P4 / molar mass P4

Moles P4 = 5.77 grams / 123.90 g/mol

Moles P4 = 0.0466 moles

Step 4: Calculate moles O2

Moles O2 = mass O2 / molar mass O2

Moles O2 = 5.77 grams / 32.0 g/mol

Moles O2 = 0.1803 moles

Step 5: Calculate limiting reactant

P4 is the limiting reactant in this reaction. It will completely be consumed (0.0466 moles). O2 is in excess, there will react 3*0.0466 = 0.1398 moles

There will remain 0.1803 - 0.1398 = 0.0405 moles O2

Step 6: Calculate the amount of P4O6

For 1 mol P4 we'll have 1 mol P4O6

For 0.0466 moles P4 we'll have 0.0466 moles P4O6

Step 7: The balanced equatio

P4O6 + 2O2 → P4O10

We have 0.0466 moles P4O6 and 0.0405 moles O2

Step 8: Calculate the limiting reactant

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

O2 is the limiting reactant. It will completely be consumed (0.0405 moles)

P4O6 is in excess. There will react 0.0405/2 = 0.02025 moles

There will remain 0.0466 - 0.02025 = 0.02635 moles P4O6

This is 0.02635 * 219.88 g/mol = 5.79 grams P4O6

Step 9: Calculate moles and mass of P4O10

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

For 0.0405 moles O2 we'll have 0.02025 moles P4O10

This is 0.02025 * 283.89 g/mol = 5.75 grams P4O10

3 0

2 years ago

What is the hydronium ion concentration of a solution with a pOH of 7.20?

-Dominant- [34]

Answer:

[H⁺] = 1.58 x 10⁻⁷ M.

Explanation:

∵ pOH = - log[OH⁻]

7.20 = - log[OH⁻]

log[OH⁻] = - 7.20

∴ [OH⁻] = 6.31 x 10⁻⁸.

∵ [H⁺][OH⁻] = 10⁻¹⁴.

∴ [H⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(6.31 x 10⁻⁸) = 1.585 x 10⁻⁷ M.

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2 years ago

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Which statement accurately compares ionic and covalent bonding?

Travka [436]

Answer:

Conduct electricity when they are molten, while covalent compounds usually do not conduct electricity when they are molten.

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2 years ago

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On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.

Delicious77 [7]

Answer:

Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes.

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

3) Now locate and label the points:

(90, 0.9) ⇒ 90 mb, 0.9 ml

(100, 0.8) ⇒ 100 mb, 0.8 ml

(400, 0.2) ⇒ 400 mb, 0.2 ml

(600, 0.15) ⇒ 600 mb, 0.15 ml

(760, 0.1) ⇒ 760 mb, 0.1 ml

The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

With that you can now locate each point on your graph.

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2 years ago

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. Divide 94.20 g by 3.167 22 mL.

jeka94

94.20 g/3.16722 mL = 29.74 g/mL

The ratio of mass to volume is equal to the substance's density. Thus, 29.74 g/mL is the density of whatever substance it may be. Density does not change for incompressible matter like solid and some liquids. Although, it may be temperature dependent.

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2 years ago

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